91Ó°ÊÓ

Density of iron, $\mathrm{ÒÏ}=7.87×{10}^3\text{kg}/{\text{m}}^3$.

Mass of iron, $m=55.934\text{u}$.

Conversion: $1\text{u}=1.66×{10}^{−27}\text{kg}$.

The radius of the nucleusis$r=A^{1/3}×R_o$ .

Where, is the mass number androle="math" localid="1658403049741" $R_0$ is role="math" localid="1658403057498" $1.2×{10}^{−15}\text{m}$.

Volume of sphere, $r=A^{1/3}×R_o$.

$V_{\text{total}}=\frac{4}{3}\mathrm{Ï€}{r}^3$

Where, $r$ is the radius

Volume of $1$ iron nucleus is given as shown below.

$\begin{array}{l}V=\frac{4}{3}\mathrm{π}{r}^3\\ V=A×\frac{4}{3}\mathrm{π}{R}_o^3\\ V=56\left(\frac{4}{3}\mathrm{π}\right){(1.2×{10}^{−15}\text{m})}^3\\ V=4.05×{10}^{−43}{\text{m}}^3\end{array}$

The number of iron atoms in a piece of solid iron of 1 cubic meter is given as:

$\begin{array}{l}N=\frac{themassof1cubicmetcrofiron}{massofoneironatom}\\ N=\frac{\mathrm{ÒÏ}×(1m^3)}{m}\\ N=\frac{(7.87×{10}^{3}kg/m^3)×(1m^3)}{(55.934u)\left(\frac{1.66×{10}^{-2k{e}_e}}{1u}\right)}\\ N=8.48×{10}^{28}iron\text{ }nucleus\end{array}$

Total volume occupied by nucleus present in $1{\text{m}}^3$.

$\begin{array}{l}{V}_{\text{total}}=N×V\\ V_{\text{total}}=(4.05×{10}^{−43}\frac{{\text{m}}^3}{1\text{nucleus}})(8.48×{10}^{28}\text{nucleus})\\ V_{\text{total}}=3.43×{10}^{−14}{\text{m}}^3\end{array}$

So, the fraction of space occupied by the iron nucleus in $1{\text{m}}^3$ is given as:

$\begin{array}{l}=\frac{V_{\text{kat1}}}{1{\text{m}}^3}\\ =\frac{3.43×{10}^{−14}{\text{m}}^3}{1{\text{m}}^3}\\ =3.43×{10}^{−14}\end{array}$

Therefore, the fraction in a piece of iron that is occupied by the iron nuclei is approximate $3×{10}^{−14}$